C++ How to put an integer into an character array

Question

I'm going mad trying to figure this out but nothing I try is working. Please help me out here.

char ch[5];
int in = 15;
ch[2] = in; // It has to be on a specefic place in that array
cout << ch[2]; // Prints '*'

I know i can do something like ch[2] = in+48; but that will only work with numbers smaller than 10.

int is bigger than char. Are you trying to truncate the int value to the size of a char, or split the int up into multiple char's? — Peter Bloomfield
– Peter Bloomfield, Commented Jan 17, 2014 at 17:15
You can't. What that does is type-cast the integer to a char, which basically chops off the high bits and only leaves the eight low bits which happens to be the same value as the ASCII code for '*'. — Some programmer dude
– Some programmer dude, Commented Jan 17, 2014 at 17:15

Joseph Mansfield · Accepted Answer · 2014-01-17 17:14:54Z

1

When cout receives a char, it interprets it as a character (according to the execution character set) rather than just an integer value. If you want to print the integer value, you'll need to cast it to some other integer type:

cout << static_cast<int>(ch[2]);

Note that you will almost certainly not be able to store all the values of an int in a char. If your char is 8 bits (which it probably is), then 255 is the maximum value you can store in it.

answered Jan 17, 2014 at 17:14

Joseph Mansfield

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4 Comments

Digital_Reality Over a year ago

@LightnessRacesinOrbit Interesting.. what exactly happened here?

Joseph Mansfield Over a year ago

@LightnessRacesinOrbit I prefer the cast because it expresses intention.

Lightness Races in Orbit Over a year ago

@Digital_Reality: Unary + results in integral promotion.

Lightness Races in Orbit Over a year ago

@sftrabbit: Me too ;)

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